# BZOJ2243 [SDOI2011]染色（树链剖分+线段树合并）

$lc$代表该区间的最左端的颜色，$rc$代表该区间的最右端的颜色

$s$代表该区间的所有连续颜色段数（仅考虑该区间）

$lazy$表示延迟信息。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define lson		i << 1, L, mid
#define rson		i << 1 | 1, mid + 1, R

typedef long long LL;

const int N = 1e5 + 10;
const int A = 21;

vector <int> v[N];
int n, m;
int c[N];
int f[N][A];
int lazy[N << 2], s[N << 2], lc[N << 2], rc[N << 2];
int h[N], deep[N], sz[N];
int son[N];
int tot = 0;
int top[N];

int LCA(int a, int b){
if (deep[a] < deep[b]) swap(a, b);
for (int i = 0,  delta = deep[a] - deep[b]; delta; delta >>= 1, ++i) if (delta & 1) a = f[a][i];
if (a == b) return a;
dec(i, 19, 0) if (f[a][i] != f[b][i]) a = f[a][i], b = f[b][i];
return f[a][0];
}

inline void pushup(int i){
lc[i] = lc[i << 1];
rc[i] = rc[i << 1 | 1];
if (rc[i << 1] ^ lc[i << 1 | 1]) s[i] = s[i << 1] + s[i << 1 | 1];
else s[i] = s[i << 1] + s[i << 1 | 1] - 1;
}

inline void pushdown(int i, int L, int R){
int tmp = lazy[i];
if (tmp == -1 || L == R) return;

s[i << 1]      = s[i << 1 | 1]    = 1;
lazy[i << 1]   = lazy[i << 1 | 1] = tmp;

lc[i << 1]     = rc[i << 1]       = tmp;
lc[i << 1 | 1] = rc[i << 1 | 1]   = tmp;
lazy[i] = -1;
}

void dfs1(int x, int fa, int dep){
sz[x] = 1;
deep[x] = dep;

if (fa){
f[x][0] = fa;
for (int i = 0; f[f[x][i]][i]; ++i) f[x][i + 1] = f[f[x][i]][i];
}

int ct = (int)v[x].size();

rep(i, 0, ct - 1){
int u = v[x][i];
if (u == fa) continue;
dfs1(u, x, dep + 1);
sz[x] += sz[u];
if (sz[son[x]] < sz[u]) son[x] = u;
}
}

void dfs2(int x, int tp){
h[x] = ++tot;
top[x] = tp;
if (son[x]) dfs2(son[x], tp);

int ct = (int)v[x].size();
rep(i, 0, ct - 1){
int u = v[x][i];
if (u == f[x][0] || u == son[x]) continue;
dfs2(u, u);
}
}

void build(int i, int L, int R){
s[i] = 1;
lazy[i] = -1;

if (L == R) return;
int mid = (L + R) >> 1;

build(lson);
build(rson);
}

void change(int i, int L, int R, int l, int r, int val){
pushdown(i, L, R);
if (L == l && R == r){
lc[i] = rc[i] = val;
s[i]  = 1;
lazy[i] = val;
return;
}

int mid = (L + R) >> 1;
if (r <= mid) change(lson, l, r, val);
else if (l > mid) change(rson, l, r, val);
else{
change(lson, l, mid, val);
change(rson, mid + 1, r, val);
}

pushup(i);
}

int query(int i, int L, int R, int l, int r){
pushdown(i, L, R);
if (L == l && R == r) return s[i];

int mid = (L + R) >> 1;
if (r <= mid) return query(lson, l, r);
else if (l > mid) return query(rson, l, r);
else{
int tmp = 1;
if (rc[i << 1] ^ lc[i << 1 | 1]) tmp = 0;
return query(lson, l, mid) + query(rson, mid + 1, r) - tmp;
}
}

int getcolor(int i, int L, int R, int x){
pushdown(i, L, R);
if (L == R) return lc[i];
int mid = (L + R) >> 1;
if (x <= mid) return getcolor(lson, x);
else return getcolor(rson, x);
}

int solvesum(int x, int tp){
int ret = 0;
for (; top[x] ^ top[tp] ;){
ret += query(1, 1, n, h[top[x]], h[x]);
if (getcolor(1, 1, n, h[top[x]]) == getcolor(1, 1, n, h[f[top[x]][0]])) --ret;
x = f[top[x]][0];
}

ret += query(1, 1, n, h[tp], h[x]);
return ret;
}

void solvechange(int x, int tp, int val){
for (; top[x] ^ top[tp]; ){
change(1, 1, n, h[top[x]], h[x], val);
x = f[top[x]][0];
}

change(1, 1, n, h[tp], h[x], val);
}

void solve(){
int x, y, z;
dfs1(1, 0, 0);
dfs2(1, 1);
build(1, 1, n);
rep(i, 1, n) change(1, 1, n, h[i], h[i], c[i]);

rep(i, 1, m){
char ch[10];
scanf("%s", ch);
if (ch[0] == 'Q'){
scanf("%d%d", &x, &y);
int t = LCA(x, y);
printf("%d\n", solvesum(x, t) + solvesum(y, t) - 1);
}
else{
scanf("%d%d%d", &x, &y, &z);
int t = LCA(x, y);
solvechange(x, t, z);
solvechange(y, t, z);
}
}
}

void init(){
scanf("%d%d", &n, &m);
rep(i, 1, n) scanf("%d", c + i);
rep(i, 2, n){
int x, y;
scanf("%d%d", &x, &y);
v[x].push_back(y);
v[y].push_back(x);
}
}

int main(){
init();
solve();
return 0;
}

posted @ 2017-07-13 14:20  cxhscst2  阅读(183)  评论(0编辑  收藏  举报